It is known that to compute the Lebesgue outer measure $\lambda^*$ of irrationals $\Bbb{I}$ we make use of the following property:
$$\lambda^*(\Bbb{I}) +\lambda^* (\Bbb{Q}) = \lambda^* (\Bbb{R})$$
and noting that since $\Bbb{Q}$ has zero measure being a countable union of singletons and using countable subadditivity of $\lambda^*$, we thus conclude that $\Bbb{I}$ has full measure and more generally $\lambda^*([a,b]\cap \Bbb{I}) = b-a$
However, in order to better understand what features of uncountable sets controls the value of its Lebesgue outer measure, I tried to compute $\lambda^*([a,b]\cap \Bbb{I})$ by constructing an open cover to exclude all rationals (Without loss of generality, and for ease of writing down the enumeration of rationals I use, we consider $\lambda^*([-1,1]\cap \Bbb{I})$ and the general result can be obtained by appropriate translation and scaling of the interval)
I first give an enumeration of the rationals within $[-1,1]$ as follows:
$S=\{(p, q) \in \Bbb{N}^2 - \{n\in \Bbb{N} : (n+2,1)\} : \pm \frac{p}{q} \in \Bbb{Q} \land \text{lcm} (p,q)=pq\}=\{0,1,-1,\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...\}$
and then construct my open cover $C$ in stages (which for the later stages, I abbreviated the union by using n-tuples). Each later stage splits up the interval from the previous stage by the next two rationals in the order given by $S$:
\begin{align} C_0 & =(-1,1)\\ C_1 & = (-1,0) \cup (0,1) := (-1,0,1)\\ C_2 & = (-1,-\frac{1}{2}) \cup (-\frac{1}{2},0) \cup (0,\frac{1}{2}) \cup (\frac{1}{2},1) := (-1,-\frac{1}{2},0,\frac{1}{2},1)\\ C_3 & = (-1,-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},1)\\ C_4 & = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},0,\frac{1}{3},\frac{1}{2},\frac{2}{3},1)\\ C_5 & = (-1,-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},1)\\ C_6 & = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},0,\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},1)\\ C_7 & = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{1}{4}, -\frac{1}{5},0,\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4},1)\\ C_8 & = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{1}{3},-\frac{2}{5}, -\frac{1}{4}, -\frac{1}{5},0,\frac{1}{5},\frac{1}{4},\frac{2}{5},\frac{1}{3},\frac{1}{2}, \frac{2}{3},\frac{3}{4},1)\\ C_9 & = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3}, -\frac{2}{5},-\frac{1}{4},-\frac{1}{5},0,\frac{1}{5},\frac{1}{4},\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3},\frac{3}{4},1)\\ C_{10} & = (-1,-\frac{3}{4},-\frac{2}{3},-\frac{1}{2},-\frac{3}{5},-\frac{1}{3}, -\frac{2}{5}, -\frac{1}{4},-\frac{1}{5},-\frac{1}{6},0,\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{2}{5},\frac{1}{3}, \frac{3}{5},\frac{1}{2}, \frac{2}{3},\frac{3}{4},1) \end{align}
and therefore: $C = \lim_{k\to \infty}C_k$
By construction, $C$ will exclude all rationals. We can also show that the lengths of all the intervals will converge to zero by considering the following subsequence $T$ of $S$
$T=\{n : \Bbb{N}-\{0\} : \pm \frac{1}{n}\} \cup \{0\} = \{-1,-\frac{1}{2},-\frac{1}{3},-\frac{1}{4},-\frac{1}{5},-\frac{1}{6},...,0,...\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{1}{2},1\}$
and the difference between consecutive terms (except $0$) given by $\frac{1}{n}-\frac{1}{n+1} = \frac{1}{n(n+1)}$ which tends to zero as $n \to \infty$, and that any elements of $S$ will lie somewhere between the elements of $T$ thus the lengths of intervals with those endpoints will be $\leq \frac{1}{n(n+1)}$ for some $n \in \Bbb{N}$
However the above conclusion result in a couple of confusing issues:
- $\Bbb{I} \cap [-1,1]$ is neither open nor closed (in the open interval topology of $\Bbb{R}$), but $C$ must exclude all rationals by construction (because $S$ enumerates all the rationals in $[-1,1]$, including troublesome limit of endpoints like $0$). This brought into question on what those progressively getting smaller intervals in $C$ are containing.
- Every interval, no matter how small, must contain uncountable many irrationals and countably many rationals. But $C$ excludes all the rationals, so that would mean those intervals contains only irrationals. But this contradicts with the fact that no uncountable subset of irrationals are open nor closed (though finite union of irrationals can be closed as finite union of singletons (which are closed sets) are closed) thus no open interval can contain only irrationals
- If the intervals tends to zero, it means their infimum length will be zero as $C$ only has countably many of them (thus by countable subadditivity of $\lambda^*$ this sums to zero). But we know this is incorrect as using the standard proofs in paragraph 1, we would conclude $\lambda^*([-1,1]\cap \Bbb{I}) = 2$ and not zero measure.
Since $C$ must exclude all rationals by construction, what exactly are the intervals in $C$ are containing in order to be consistent with the issues raised in 1,2,3? Does $C$ actually contains intervals?